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SOLVABILITY CONDITIONS FOR SOME DIFFERENCE OPERATORS N. C. APREUTESEI AND V. A. VOLPERT Received 24 June 2004 Infinite-dimensional difference operators are studied. Under the assumption that the coefficients of the operator have limits at infinity, limiting operators and associated polynomials are introduced. Under some specific conditions on the polynomials, the operator is Fredholm and has the zero index. Solvability conditions are obtained and the exponential behavior of solutions of the homogeneous equation at infinity is proved. 1. Introduction Infinite-dimensional difference operators may not satisfy the Fredholm property, and the Fredholm-type solvability conditions are not necessarily applicable to them. In other words, we do not know how to solve linear algebraic systems with infinite matrices. Various properties of linear and nonlinear infinite discrete systems are studied in [1, 2, 3, 4, 5, 6, 7, 8]. The goal of this paper is to establish the normal solvability for the difference operators of the form j j j (Lu) j = a−m u j −m + · · · + a0 u j + · · · + am u j+m , j ∈ Z, (1.1) and to obtain the solvability conditions for the equation Lu = f , where m ≥ 0 is a given integer and f = { f j }∞ j =−∞ is an element of the Banach space
     j =−∞ , u j ∈ R, sup u j < ∞ .  ∞ E = u = uj j ∈Z (1.2) The right-hand side in (1.1) does not necessarily contain an odd number of summands. We use this form of the operator to simplify the presentation. We will use here the approaches developed for elliptic problems in unbounded domains [9, 10] and adapt them for infinite-dimensional difference operators. Copyright © 2005 Hindawi Publishing Corporation Advances in Difference Equations 2005:1 (2005) 1–13 DOI: 10.1155/ADE.2005.1 2 Solvability conditions for some difference operators The operator L : E → E defined in (1.1) can be regarded as (Lu) j = A j U j , where  j j j  A j = a−m ,...,a0 ,...,am ,  U j = u j −m ,...,u j ,...,u j+m  (1.3) are 2m + 1-vectors, A j is known, and U j is variable. We suppose that there exist the limits of the coefficients of the operator L as j → ±∞ j a±l = lim al , j →±∞ l ∈ Z, −m ≤ l ≤ m, (1.4) and a±±m = 0. Denote by L± : E → E the limiting operators   L± u j = a±−m u j −m + · · · + a±0 u j + · · · + a±m u j+m , j ∈ Z. (1.5) Recall that a linear operator L : E → E is normally solvable if its image Im L is closed. If L is normally solvable with a finite-dimensional kernel and the codimension of its image is also finite, then L is called Fredholm operator. Denoting by α(L) and β(L) the dimension of ker L and the codimension of ImL, respectively, we can define the index κ(L) of the operator L as κ(L) = α(L) − β(L). It is known that the index does not change under deformation in the class of Fredholm operators. In Section 2 of this paper we introduce polynomials P + (σ) and P − (σ) associated with the limiting operators L+ and L− . We show that, if P + and P − do not have roots on the unit circle, then the limiting operators are invertible and the operator L is normally solvable with a finite-dimensional kernel. If moreover the polynomials have the same number of roots inside the unit circle, then L is a Fredholm operator and its index is zero. In Section 3 we prove that under some conditions on the polynomials P + and P − corresponding to operator L in (1.1), the bounded solutions of the equation Lu = 0 are exponentially decreasing at +∞ and −∞. The idea is to approximate the equation Lu = 0 at +∞ with the problem on half-axis:   L+ u j = 0, j ≥ 1, (1.6) u1 = a1 ,...,uk = ak , and similarly at −∞. We first prove that (1.6) has a unique solution and that this solution is exponentially decaying. Then we deduce that the solution of the equation Lu = 0 is also exponentially decaying. Section 4 deals with the solvability conditions for the equation Lu = f , for L in (1.1) and f = { f j }∞ j =−∞ ∈ E being given. Let L∗ be the formally adjoint of L, α(L∗ ) = dim(kerL∗ ), and vl = {vlj }∞ j =−∞ , l = 1,...,α(L∗ ) some linearly independent solutions of the equation L∗ v = 0. One states a result which is analogous to the continuous case: equation Lu = f is solvable if and only if f is orthogonal on all solutions vl , l = 1,...,α(L∗ ). Section 5 is devoted to a particular case of the operator L related to discretization of a second-order differential equation on the real axis:   (Lu) j = u j+1 − 2u j + u j −1 + c j u j+1 − u j + b j u j , j ∈ Z, (1.7) N. C. Apreutesei and V. A. Volpert 3 ∞ where {b j }∞ j =−∞ and {c j } j =−∞ are given sequences of real numbers. If there exist the lim± its b = lim j →±∞ b j < 0, and c± = lim j →±∞ c j ≥ −2, then there are no roots of the polynomials on the unit circle, and there is exactly one root inside it. Therefore the bounded solution of the equation Lv = 0 is exponential decaying at ±∞. If moreover c± ∈ [−2,2], then the solvability conditions for the equation Lu = f are applicable. 2. Limiting operators and normal solvability Let E be the Banach space of all bounded real sequences E = {u = {u j }∞ j =−∞ , u j ∈ R, sup j ∈Z |u j | < ∞} with the norm   u = sup u j , (2.1) j ∈Z j j j ∈ Z, (2.2) and let L : E → E be the general linear difference operator (Lu) j = a−m u j −m + · · · + a0 u j + j j j j · · · + am u j+m , j ∈ Z, where m ≥ 0 is an integer and a−m ,...,a0 ,... ,am ∈ C are given coefficients. Denote by L+ : E → E the limiting operator   L+ u j = a+−m u j −m + · · · + a+0 u j + · · · + a+m u j+m , where j a+l = lim al , j →∞ l ∈ Z, −m ≤ l ≤ m. (2.3) We are going to define the associated polynomial for the operator L+ . To do this, we are looking for the solution of the equation L+ u = 0 under the form u j = exp(µ j), j ∈ Z, and obtain a+−m e−µm + · · · + a+−1 e−µ + a+0 + a+1 eµ + · · · + a+m eµm = 0. (2.4) One takes σ = eµ and finds the polynomial associated to L+ : P + (σ) = a+m σ 2m + · · · + a+0 σ m + · · · + a+−m . (2.5) Recall the following auxiliary result from [1]. Lemma 2.1. The equation L+ u = 0 has nonzero bounded solutions if and only if the corresponding algebraic polynomial P + has a root σ with |σ | = 1. We will find conditions in terms of P + for the limiting operator L+ to be invertible. One begins with an auxiliary result concerning continuous deformations of the polynomial P + . Without loss of generality, we may assume that the coefficient a+m = 1. Consider the polynomial with complex coefficients P(σ) = σ n + a1 σ n−1 + · · · + an−1 σ + an . (2.6) Lemma 2.2. Suppose that a polynomial P(σ) does not have roots with |σ | = 1 and it has k roots with |σ | < 1, 0 ≤ k ≤ n. Then there exists a continuous deformation Pτ (σ), 0 ≤ τ ≤ 1, 4 Solvability conditions for some difference operators such that    P1 (σ) = σ k − a σ n−k − λ , P0 (σ) = P(σ), (2.7) and the polynomial Pτ (σ) does not have roots with |σ | = 1 for any 0 ≤ τ ≤ 1. Here λ > 1 and a < 1 are real numbers. Proof. We represent the polynomial P(σ) in the form     P(σ) = σ − σ1 · · · σ − σn , (2.8) where the roots σ1 ,...,σk are inside the unit circle, and the other roots are outside it. Consider the polynomial    Pτ (σ) = σ − σ1 (τ) · · · σ − σn (τ)  (2.9) that depends on the parameter τ through its roots. This means that we change the roots and find the coefficients of the polynomial through them. We change the roots in such a way that for τ = 0 they coincide with the roots of the original polynomial; for τ = 1 it has the roots σ1 ,...,σk with (σi )k = a, i = 1,...,k (inside the unit circle) and n − k roots σk+1 ,...,σn such that (σi )n−k = λ, i = k + 1,...,n (outside of the unit circle). This deformation can be done in such a way that there are no roots with |σ | = 1. The lemma
is proved. Using the associated polynomials P + and P − of L+ and L− , we can study the normal solvability of the operator L. Theorem 2.3. The operator L is normally solvable with a finite-dimensional kernel if and only if the corresponding algebraic polynomials P + and P − do not have roots σ with |σ | = 1. Proof The necessity. Suppose that the polynomials P + , P − do not have roots σ with |σ | = 1. We first show that the image of L is closed. To do this, let { f n } be a sequence in ImL such that f n → f and let {un } be a sequence with the property Lun = f n . Suppose in the beginning that {un } is bounded in E. We construct a convergent subsequence. Since un  = sup j ∈Z |unj | ≤ c, then for every positive integer N, there exists a subsequence {unk } of {un } and an element u = {u j }Nj=−N ∈ E such that   sup unj k − u j  −→ 0, −N ≤ j ≤N (2.10) that is, unk → u as k → ∞ uniformly on each bounded interval of j. Using a diagonalization process, we extend u j to all j ∈ Z. It is clear that sup j ∈Z |u j | ≤ c; that means u ∈ E. Passing to the limit as k → ∞ in the linear equation Lunk = f nk , we get Lu = f , so f ∈ ImL. N. C. Apreutesei and V. A. Volpert 5 We show that the convergence in (2.10) is uniform with respect to all j ∈ Z. Supposing by contradiction that there exists jk → ∞ such that |unjkk − u jk | ≥ ε > 0, observe that the nk nk k k sequence y k = { y kj }∞ j =−∞ , y j = u j+ jk − u j+ jk verifies the inequality | y0 | = |u jk − u jk | ≥ ε and the equation j+ j j+ jk k yj a−mk y kj−m + · · · + a0 j+ j nk + · · · + am k y kj+m = f j+ jk − f j+ jk , j ∈ Z. (2.11) Since the sequence { y k } is bounded in E, there exists a subsequence { y kl } which converges (say y kl → y 0 ) uniformly with respect to j on bounded intervals. We may pass to the limit as kl → ∞ in (2.11) and obtain via (2.3), a+−m y 0j −m + · · · + a+0 y 0j + · · · + a+m y 0j+m = 0, j ∈ Z. (2.12) Thus, the limiting equation L+ u = 0 has a nonzero bounded solution y 0 = { y 0j }∞ j =−∞ . Lemma 2.1 leads to a contradiction. Therefore the convergence unj k − u j → 0 is uniform with respect to all j ∈ Z and, since Lu = f , it follows that Im L is closed. We analyze now the case when {un } is unbounded in E. Then we write un = xn + y n with {xn } ∈ kerL and { y n } is in the supplement of kerL. Then Ly n = f n . If { y n } is bounded in E, it follows as above that Im L is closed. If not, then we repeat the above reasoning for zn = y n /  y n  and g n = f n /  y n . Passing to the limit on a subsequence nk (such that znk → z0 ) in the equality Lznk = g nk and using the convergence g nk → 0, one obtains the contradiction that z0 ∈ kerL. Therefore Im L is closed. In order to prove that kerL has a finite dimension, it suffices to show that every sequence un from B ∩ kerL (where B is the unit ball) has a convergent subsequence. The reasoning is similar to that of the first part, taking f n = 0. The sufficiency. Assume that ImL is closed and dim(kerL) is finite. By contradiction, one supposes that either P + or P − (say P + ) has a root on the unit circle. Then the correspondiξ j ing solution of L+ u = 0 has the form u = {u j }∞ j =−∞ , where u j = e , ξ ∈ R, j ∈ Z. N ∞ N ∞ N N N Let α = {α j }∞ j =−∞ , β = {β j } j =−∞ , γ = {γ j } j =−∞ be a partition of unity (α j + β j + N γ j = 1) given by  1, αj =  0, j ≤ 0, j ≥ 1, βNj =  0, 1 ≤ j ≤ N, j ≤ 0, j ≥ N + 1, γNj j ≥ N + 1, j ≤ N.  1, =  1, 0, (2.13) n ∞ n ∞ n n For a fixed εn → 0 (as n → ∞), let un = {unj }∞ j =−∞ , v = {v j } j =−∞ , f = { f j } j =−∞ be the n n n n n i(ξ+ε ) j n sequences defined by u j = e , v j = (1 − α j )(u j − u j ), and f j = Lv j , j ∈ Z. It is clear that unj → u j (n → ∞) uniformly on every bounded interval of integers j. 6 vn Solvability conditions for some difference operators It is sufficient to prove that f n → 0. Indeed, in this case, by hypothesis it follows that → 0. But this is in contradiction with   vn = sup ei(ξ+εn ) j − eiξ j  ≥ m > 0 (2.14) j>0 for some m. In order to show that f n → 0 as n → ∞, observe that f jn can be written under the form     f jn = α j + βNj + γNj L βN + γN un − u   j   N   N = αj L β + γ u − u j + β j L β + γN un − u          + γNj L βN un − u j + γNj L − L+ γN un − u j    N N     n + γNj L+ γN un − u    j (2.15) j. A simple computation implies that the first three terms tend to zero as n → ∞, uniformly with respect to all integers j. Next, condition (2.3) and the boundedness un  = u = 1 lead to the convergence  N          γ L − L+ γN un − u j  ≤ γN L − L+ 0 · γN (un − u) −→ 0, j (2.16) as N → ∞, where | · |0 is the norm of the operator. For a given N, one estimates the last term of (2.15). Since u j = eiξ j , j ∈ Z is a solution of the equation L+ u = 0, then   L + un − u  j       = L+ un j = L+ un j − eiεn j L+ u j     = ei(ξ+εn ) j a+−m e−iξm e−iεn m − 1 + · · · + a+−1 e−iξ e−iεn − 1     + a+1 eiξ eiεn − 1 + · · · + a+m eiξm eiεn m − 1 , (2.17) so   L + un − u  j = iεn ei(ξ+εn ) j a+−m (−m)e−iξm eib−m + · · · − a+−1 e−iξ eib−1 + a+1 eiξ eib1 + · · · + a+m meiξm eibm , j ∈ Z, (2.18) where b−m ,...,b−1 ,b1 ,...,bm are intermediate points. Thus the last term in (2.15) goes to zero as n → ∞ and therefore f n → 0. This completes the proof.
Now we are ready to establish the invertibility of L+ . Theorem 2.4. If the operator L+ is such that the corresponding polynomial does not have roots with |σ | = 1, then it is invertible. N. C. Apreutesei and V. A. Volpert 7 Proof. Lemma 2.2 for P + implies the existence of a continuous deformation Pτ (σ), 0 ≤ τ ≤ 1, from the polynomial P0 = P + to P1 (σ) = (σ k − a)(σ 2m−k − λ) such that Pτ (σ) does not admit solutions with |σ | = 1. Here λ > 1, a < 1 are given. The operator which corresponds to P1 is L+1 defined by   L+1 u j = u j+k − au j − λu j+2k−2m + aλu j+k−2m . (2.19) Indeed, looking for the solution of L+1 in the form u j = eµ j , we arrive at eµk − a − λeµ(2k−2m) + aλeµ(k−2m) = 0. (2.20) We put σ = eµ and get    σ k − a σ 2m−k − λ = 0, (2.21) so P1 is the above polynomial. Taking a = 1/λ, we obtain   1 L+1 u j = (Mu) j − u j , λ (2.22) (Mu) j = u j+k − λu j+2k−m + u j+k−2m (2.23) where is invertible for large λ ≥ 0 (see, e.g., [1, Lemma 4.9]). Since L+1 is close to M (for λ ≥ 0 large enough), one deduces that L+1 is also invertible. Hence the index of L+1 is zero. Since the continuous deformation Pτ does not have solutions σ with |σ | = 1, we find that the corresponding continuous deformation of the operator L+τ does not admit nonzero bounded solutions (see Lemma 2.1). By Theorem 2.3 one obtains that L+τ is normally solvable with a finite-dimensional kernel. From the general theory of Fredholm operators, we know that the index of such homotopies does not change. Since the index of L+1 is κ(L+1 ) = 0, we deduce that κ(L+ ) = 0. This, together with the fact that ker L+ = Φ, implies ImL+ = E, therefore L+ is invertible. The theorem is proved.
Remark 2.5. An analogous result can be stated for L− . As a consequence, we may study the Fredholm property of L with the aid of the polynomials P + and P − . Corollary 2.6. If the limiting operators L+ and L− for an operator L are such that the corresponding polynomials P + (σ) and P − (σ) do not have roots with |σ | = 1 and have the same number of roots inside the unit circle, then L is a Fredholm operator with the zero index. Proof. We construct a homotopy of L in such a way that L+ and L− are reduced independently to the operator in Theorem 2.4. Then, this homotopy is in the class of the normally solvable operators with finite-dimensional kernels. Since at +∞ and −∞ the operators L+ and L− coincide, we finally reduce L to an operator with constant coefficients. According to Theorem 2.4, it is invertible. Therefore, L is a Fredholm operator and has the zero index, as claimed.
8 Solvability conditions for some difference operators 3. Exponential decay We consider now the problem   L+ u j = 0, j ≥ 1, (3.1) assuming that the corresponding polynomial P + (σ) does not have roots with |σ | = 1 and has k roots with |σ | < 1. One associates to (3.1) the boundary conditions u1 = a1 ,...,uk = ak , (3.2) with a1 ,...,ak ∈ R given. Since there are k roots inside the unit circle, then there are k linearly independent solutions of the equation L+ u = 0 decaying as j → ∞. Denote them by u1 ,...,uk . Consider their values for j = 1,...,2m: u11 ,u12 ,...,u12m , .. . (3.3) uk1 ,uk2 ,...,uk2m . Each of these solutions is completely determined by the above values. Therefore the corresponding k vectors are linearly independent. Indeed, otherwise the solutions would have been linearly dependent. Therefore there exist k linearly independent columns. Without loss of generality we can assume that these are the first k columns. Hence the corresponding k × k matrix is invertible. Any bounded solution of (3.1) can be represented in the form u = c1 u1 + · · · + ck uk . (3.4) Substituting it in (3.2), we uniquely determine the coefficients c1 ,...,ck . Therefore we have proved the following result. Proposition 3.1. If the corresponding polynomial P + (σ) for L+ does not admit roots on the unit circle |σ | = 1 and has k roots with |σ | < 1, then for each (a1 ,...,ak ) ∈ Rk , problem (3.1)-(3.2) has a unique bounded solution. In addition, this solution is exponentially decreasing. This result holds also for L− . Thus, for the solution of Lu = 0, we may conclude the following. Theorem 3.2. Suppose that the polynomials P + (σ) and P − (σ) corresponding to L+ and L− , respectively, do not have roots with |σ | = 1 and have the same number of roots with |σ | < 1. Then the bounded solutions of the equation Lu = 0 are exponentially decreasing at ±∞. Proof. Let ũ = {ũ j }∞ j =−∞ be a bounded solution of the equation Lu = 0. Consider the problem (Lu) j = 0, uN+1 = a1 ,...,uN+k = ak . (3.5) N. C. Apreutesei and V. A. Volpert 9 For N sufficiently large this problem is uniquely solvable for any a1 ,...,ak since problem (3.1)-(3.2) is uniquely solvable, and the operator L is close to the operator L+ . If we put ai = ũN+i , i = 1,...,k, then the solution of problem (3.5) coincides with ũ for j ≥ N. Therefore it is sufficient to prove that the solution of problem (3.5) is exponentially decreasing for any ai , i = 1,...,k.  Consider the operator S of multiplication by exp(µ 1 + j 2 ), that is, (Su) j = eµ √ 1+ j 2 uj, j = 0, ±1,.... (3.6) Here µ > 0. Let Lµ = SLS−1 . Then   j Lµ u j = a−m u j −m eµ( √ √ 1+ j 2 − 1+( j −m)2 ) j j + · · · + a0 u j + · · · + am u j+m eµ( √ 1+ j 2 − √ 1+( j+m)2 ) . (3.7) For µ sufficiently small, the operator Lµ is close to the operator L. Therefore the problem  Lµ v  j = 0, vN+1 = b1 ,...,vN+k = bk (3.8) is uniquely solvable for any b1 ,...,bk . If we put bi = exp(µ 1 + (N + i)2 )ai , i = 1,...,k, then the solution u of problem (3.5) can be expressed through the solution v of problem (3.8): u = S−1 v. Since v is bounded, then u is exponentially decreasing. Thus we have proved that ũ is exponentially decreasing as j → ∞. Similarly it can be proved for j → −∞. The theorem is proved.
4. Solvability conditions In this section, we establish solvability conditions for the equation Lu = f . (4.1) Here L is the operator in (1.1) and f = { f j }∞ j =−∞ is fixed in E. For the operator L, denote α(L) = dim(kerL) and β(L) = codim(ImL). If (u,v) is the ∞ 2 inner product of two sequences u = {u j }∞ j =−∞ , v = {v j } j =−∞ in the sense l , that is, (u,v) = ∞  ujvj, (4.2) j =−∞ then we may define the formally adjoint L∗ of the operator L by the equality   (Lu,v) = u,L∗ v . (4.3) Let L+ ,L− and L+∗ ,L−∗ be the limiting operators associated with L and L∗ , respectively. We work under the following hypothesis: (H) the polynomials P + ,P − corresponding to L+ and L− do not have roots with |σ | = 1 and have the same number of roots with |σ | < 1. Similarly for the polynomials P∗+ and P∗− corresponding to L+∗ and L−∗ . 10 Solvability conditions for some difference operators Corollary 2.6 implies that L and L∗ are Fredholm operators with the zero index. Lemma 4.1. Under hypothesis (H), it holds that β(L) ≥ α(L∗ ). Proof. By the definition of the Fredholm operator it follows that (4.1) is solvable for a given f = { f j }∞ j =−∞ ∈ E if and only if there exist linearly independent functionals ϕk ∈ E∗ , k = 1,...,β(L) such that ϕk ( f ) = 0, k = 1,...,β(L). (4.4) On the other hand consider the functionals ψl given by ψl ( f ) = ∞  j =−∞ f j vlj ,   l = 1,...,α L∗ , (4.5) ∗ where vl = {vlj }∞ j =−∞ , l = 1,...,α(L ) are linearly independent solutions of the homoge∗ neous equation L v = 0. We know from Theorem 3.2 that the values vlj are exponentially decreasing with respect to j. Therefore the functionals ψl are well defined. Obviously, ψl is linear for each l. If f (n) → f in E (in the norm supremum), then we may pass to the limit in (4.5) under the sum to find that ψl ( f (n) ) → ψl ( f ), as n → ∞, for all l = 1,...,α(L∗ ), that is, ψl are continuous. Therefore ψl ∈ E∗ , l = 1,...,α(L∗ ), where E∗ denotes the dual space of E. In order to prove that β(L) ≥ α(L∗ ), suppose that it is not true. Then among the functionals ψl there exists at least one functional (say ψ1 ) which is linearly independent with respect to all ϕk , k = 1,...,β(L). This means that there exists f ∈ E such that (4.4) holds, but ψ1 ( f ) = ∞  j =−∞ f j v1j = 0. (4.6) From (4.4) it follows that (4.1) is solvable. We multiply it by v1 and find (Lu,v1 ) = ( f ,v1 ). By (4.6) observe that the right-hand side is different from zero. But since v1 is a solution of the equation L∗ v = 0, we deduce that (Lu,v1 ) = (u,L∗ v1 ) = 0. The contradiction we arrive at, proves the lemma.
Remark 4.2. Analogously we find β(L∗ ) ≥ α(L). Therefore, if one denotes by κ(L) = α(L) − β(L) the index of the operator L, we get   κ(L) + κ L∗ ≤ 0. (4.7) Since in our case κ(L) = κ(L∗ ) = 0, it follows that   β(L) = α L∗ ,   β L∗ = α(L). (4.8)